This looks a lot like physics, but it is actually a math question! I will be omitting unnecessary constants for simplicity so the units might be off.

I want to reduce the equation

$-i\omega \vec{j}(\vec{q},\omega)=\vec{E}(\vec{q},\omega) +i\vec{q}\delta n(\vec{q},\omega)$

to an equation of the following form:


with current density $\vec{j}$, conductivity tensor $\underline{\sigma}$, $\vec{E}$ the electric field and wave vector $\vec{q}$. The factors $-i\omega$ and $i\vec{q}$ come frome the derivatives of the respective fourier transforms (once with respect to time and once with respect to $\vec{x}$).

It is given that $\delta n(\vec{q},\omega)=-\vec{q}\cdot \vec{j}\frac{1}{e\omega}$ where $e$ stands for the electron charge.

Now here comes the tricky part: In the solutions they plug in the given definition of $\delta n$ and then write the last term of the first equation as

$-i\frac{1}{e\omega}\vec{q}\vec{q}^T \cdot \vec{j}$ where $\vec{q}\vec{q}^T$ produces a matrix. My question is: where does this come from? Of course a matrix has to come out eventually (as I am explicitly looking for the conductivity tensor) but in my naivety I did the following:

$i\vec{q}\delta n(\vec{q},\omega)\rightarrow-i\frac{1}{e\omega }q^2 \cdot \vec{j}$

which I could also represent as a matrix multiplication

$-i\frac{1}{e\omega }q^2 \cdot \vec{j}=-i\frac{1}{e\omega }q^2 \mathbb{I} \vec{j}$

where $\mathbb{I}$ is supposed to stand for the identity matrix.

So, when multiplicating two vectors, when do I tensor and when do I not?

  • $\begingroup$ In the equation for $\delta n$, is that a dot product on the right-hand side? $\endgroup$ Commented Mar 26, 2016 at 13:05
  • $\begingroup$ Yes. $\delta n$ is a scalar. $\endgroup$
    – AlphaOmega
    Commented Mar 26, 2016 at 14:21

1 Answer 1


If $x$ and $y$ are vectors, i.e., column matrices, and if $x^{T}$ denotes the transpose of $x$ (a row matrix), then the dot product (or inner product) may be viewed as the matrix product $$ x \cdot y = x^{T}y, $$ while the tensor product (or outer product) is $$ x \otimes y = x y^{T}. $$ (In case it matters, the matrix product $xy$ isn't defined; the sizes of the factors aren't suitable. Particularly, $q^{2}$ has no standard meaning if $q$ is a column.)

In your case (omitting the arrows and arguments), it appears the outer product of the $q$'s comes from the matrix product $$ q\, \delta n = q (q \cdot j) = q (q^{T} j) = (q q^{T}) j. $$

  • 1
    $\begingroup$ Ehrm yes, that must be it. Amazingly simple, yet I didn't see it myself. Thanks :) $\endgroup$
    – AlphaOmega
    Commented Mar 26, 2016 at 14:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .